What is the volume of 1.9 moles of chlorine gas (cl2) at 298 k and 1.3 atm?

Considering ideal gas behavior, the volume of 1 mol of gas at STP is 22.4 L; then the volume occupied by 1.9 moles is 1.9mol*22.4L/mol

The answer is 43 L if I am correct.

Hello!

For this exercise you have to use the Ideal Gas Law Equation, as follow:

PV= nRT

Where:

P= pressure ( in this example, atmosphere (atm) is the pressure unit)

V= volume (which is the value you want to calculate)

n= Mol (as reference to the mass of the gas Cl2)

R = Ideal Gas Constant (for this example is a number 0.0821 atm*L/mol*K)

T= Temperature in Kelvin (K)

So, for find the value of the volume, you just have to replace the values given in the equation as follow:

P= 1.3 atm

V = you don't know, is what you want to know.

n= 1.9 mol

R =0.0821 atmL/molK

T =298 K

Remember, never forget to put the units, because teacher can get down your score if you don't do it. So, back with the values to the equation:

13atm * V = 1.9mol * 0.0821 atmL/molK * 298K

Then, we have to solve the equation to find V value, so we has to send 1.3 atm to divide in the other side of the equation:

V = (1.9mol * 0.0821 atmL/molK * 298K)/ 1.3 atm

V = 35.76 L

And, that's it!

I hope this help you a lot, and good luck.

chemical digestion occurs in the stomach were hcl and pepsin forms chyme, the mouth in which salivary amylase breaks down starch to form bolus, and the small intestine specifically the duodenum.